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3 years ago

11

What additional info do you need

1 answer:

Novay_Z [31]3 years ago

7 0

Answer:

∠C ≅ ∠M or ∠B ≅ ∠L

Step-by-step explanation:

You are given an angle and its opposite side as being congruent. AAS requires two congruent angles and one side, so you need another set of congruent angles (one in each triangle). It does not matter which they are. The above-listed pairs are appropriate.*

_____

* Since the figure cannot be assumed to be drawn to scale, either of angles B or C could be declared congruent to either of angles L or M. However, it appears that angles B and L are opposite the longest side of the triangle, so it makes good sense to declare that pair congruent. The same congruence statement (ΔBCD≅ΔLMN) would result from declaring angles C and M congruent. So, either declaration will work (matches the last answer choice.)

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AAS requires two angles and a side. One side is already marked, so we do not need any more information about sides. (The second and third answer choices can be rejected as irrelevant.)

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2 years ago

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Please help I don’t understand please

irinina [24]

Answer:

7.2

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thats the answer

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3 years ago

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Evaluate sin^2Θ, for Θ=45°

Zielflug [23.3K]

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Step-by-step explanation:

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3 years ago

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Surface area help plz

lana [24]

Answer:

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Step-by-step explanation:

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3 years ago

Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.

likoan [24]

Answer:

Thus, the two root of the given quadratic equation $x^2+4=6x$ is 5.24 and 0.76 .

Step-by-step explanation:

Consider, the given Quadratic equation, $x^2+4=6x$

This can be written as , $x^2-6x+4=0$

We have to solve using quadratic formula,

For a given quadratic equation $ax^2+bx+c=0$ we can find roots using,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ...........(1)

Where, $\sqrt{b^2-4ac}$ is the discriminant.

Here, a = 1 , b = -6 , c = 4

Substitute in (1) , we get,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\Rightarrow x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot 1 \cdot (4)}}{2 \cdot 1}$

$\Rightarrow x=\frac{6\pm\sqrt{20}}{2}$

$\Rightarrow x=\frac{6\pm 2\sqrt{5}}{2}$

$\Rightarrow x={3\pm \sqrt{5}}$

$\Rightarrow x_1={3+\sqrt{5}}$ and $\Rightarrow x_2={3-\sqrt{5}}$

We know $\sqrt{5}=2.23607$ (approx)

Substitute, we get,

$\Rightarrow x_1={3+2.23607}$ (approx) and $\Rightarrow x_2={3-2.23607}$ (approx)

$\Rightarrow x_1={5.23607}$ (approx) and $\Rightarrow x_2=0.76393}$ (approx)

Thus, the two root of the given quadratic equation $x^2+4=6x$ is 5.24 and 0.76 .

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3 years ago

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